Sủa đề : Giải phương trình \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}+1\)
ĐKXĐ : \(x\ge1\)
\(pt\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Ta thấy : \(VT=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\Rightarrow5\le x\le10\)(TM ĐKXĐ)
Vậy \(5\le x\le10\)
